General Chemistry Principles and Modern Applications 8ed

1. Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th
Edition
Chapter 18: Additional Aspects of
Acid-Base Equilibria

2. Prentice-Hall General Chemistry: Chapter 18Slide 2 of 42
Contents
18-1 The Common-Ion Effect in Acid-Base Equilibria
18-2 Buffer Solutions
18-3 Acid-Base Indicators
18-4 Neutralization Reactions and Titration Curves
18-5 Solutions of Salts of Polyprotic Acids
18-6 Acid-Base Equilibrium Calculations: A Summary
Focus On Buffers in Blood

3. Prentice-Hall General Chemistry: Chapter 18Slide 3 of 42
18-1 The Common-Ion Effect in Acid-
Base Equilibria
• The Common-Ion Effect describes the effect on an
equilibrium by a second substance that furnishes ions
that can participate in that equilibrium.
• The added ions are said to be common to the
equilibrium.

4. Prentice-Hall General Chemistry: Chapter 18Slide 4 of 42
Solutions of Weak Acids and Strong Acids
• Consider a solution that contains both
0.100 M CH3CO2H and 0.100 M HCl.
CH3CO2H + H2O  CH3CO2
-
+ H3O+
HCl + H2O  Cl-
+ H3O+
(0.100-x) M x M x M
0.100 M 0.100 M
[H3O+
] = (0.100 + x) M essentially all due to HCl

5. Prentice-Hall General Chemistry: Chapter 18Slide 5 of 42
Acetic Acid and Hydrochloric Acid
0.1 M HCl 0.1 M CH3CO2H 0.1 M HCl +
0.1 M CH3CO2H

6. Prentice-Hall General Chemistry: Chapter 18Slide 6 of 42
Example 18-1
Demonstrating the Common-Ion Effect:
A Solution of a weak Acid and a Strong Acid.
(a) Determine [H3O+
] and [CH3CO2
-
] in 0.100 M CH3CO2H.
(b) Then determine these same quantities in a solution that is
0.100 M in both CH3CO2H and HCl.
CH3CO2H + H2O → H3O+
+ CH3CO2
-
Recall Example 17-6 (p 680):
[H3O+
] = [CH3CO2
-
] = 1.310-3
M

7. Prentice-Hall General Chemistry: Chapter 18Slide 7 of 42
Example 18-1
CH3CO2H + H2O → H3O+
+ CH3CO2
-
Initial concs.
weak acid 0.100 M 0 M 0 M
strong acid 0 M 0.100 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x 0.100 + x  0.100 M

8. Prentice-Hall General Chemistry: Chapter 18Slide 8 of 42
Example 18-1
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x 0.100 + x  0.100 M
CH3CO2H + H2O → H3O+
+ CH3CO2
-
[H3O+
] [CH3CO2
-
]
[C3CO2H]
Ka=
x · (0.100 + x)
(0.100 - x)
=
x · (0.100)
(0.100)
= = 1.810-5
[CH3CO2
-
] = 1.810-5
M compared to 1.310-3
M.
Le Chatellier’s Principle

9. Prentice-Hall General Chemistry: Chapter 18Slide 9 of 42
Suppression of Ionization
of a Weak Acid

10. Prentice-Hall General Chemistry: Chapter 18Slide 10 of 42
Suppression of Ionization
of a Weak Base

11. Prentice-Hall General Chemistry: Chapter 18Slide 11 of 42
Solutions of Weak Acids and Their Salts

12. Prentice-Hall General Chemistry: Chapter 18Slide 12 of 42
Solutions of Weak Bases and Their Salts

13. Prentice-Hall General Chemistry: Chapter 18Slide 13 of 42
18-2 Buffer Solutions
• Two component systems that change pH only
slightly on addition of acid or base.
– The two components must not neutralize each other but
must neutralize strong acids and bases.
• A weak acid and it’s conjugate base.
• A weak base and it’s conjugate acid

14. Prentice-Hall General Chemistry: Chapter 18Slide 14 of 42
Buffer Solutions
• Consider [CH3CO2H] = [CH3CO2
-
] in a solution.
[H3O+
] [CH3CO2
-
]
[C3CO2H]
Ka= = 1.810-5
= 1.810-5
[CH3CO2
-
]
[C3CO2H]
Ka[H3O+
] =
pH = -log[H3O+
] = -logKa = -log(1.810-5
) = 4.74

15. Prentice-Hall General Chemistry: Chapter 18Slide 15 of 42
How A Buffer Works

16. Prentice-Hall General Chemistry: Chapter 18Slide 16 of 42
The Henderson-Hasselbalch Equation
• A variation of the ionization constant expression.
• Consider a hypothetical weak acid, HA, and its
salt NaA:
HA + H2O  A-
+ H3O+
[H3O+
] [A-
]
[HA]
Ka=
[H3O+
]
[HA]
Ka=
[A-
]
-log[H3O+
]-log
[HA]
-logKa=
[A-
]

17. Prentice-Hall General Chemistry: Chapter 18Slide 17 of 42
Henderson-Hasselbalch Equation
-log[H3O+
] - log
[HA]
-logKa=
[A-
]
pH - log
[HA]
pKa =
[A-
]
pKa + log
[HA]
pH =
[A-
]
pKa + log
[acid]
pH =
[conjugate base]

18. Prentice-Hall General Chemistry: Chapter 18Slide 18 of 42
Henderson-Hasselbalch Equation
• Only useful when you can use initial concentrations
of acid and salt.
– This limits the validity of the equation.
• Limits can be met by:
0.1 <
[HA]
< 10
[A-
]
[A-
] > 10Ka and [HA] > 10Ka
pKa + log
[acid]
pH=
[conjugate base]

19. Prentice-Hall General Chemistry: Chapter 18Slide 19 of 42
Example 18-5
Preparing a Buffer Solution of a Desired pH.
What mass of NaC2H3O2 must be dissolved in 0.300 L of
0.25 M HC2H3O2 to produce a solution with pH = 5.09?
(Assume that the solution volume is constant at 0.300 L)
HC2H3O2 + H2O  C2H3O2
-
+ H3O+
Equilibrium expression:
[H3O+
]
[HC2H3O2]
Ka=
[C2H3O2
-
]
= 1.810-5

20. Prentice-Hall General Chemistry: Chapter 18Slide 20 of 42
Example 18-5
[H3O+
]
[HC2H3O2]
Ka=
[C2H3O2
-
]
= 1.810-5
[H3O+
] = 10-5.09
= 8.110-6
[HC2H3O2] = 0.25 M
Solve for [C2H3O2
-
]
[H3O+
]
[HC2H3O2]
= Ka
[C2H3O2
-
] = 0.56 M
8.110-6
0.25
= 1.810-5

21. Prentice-Hall General Chemistry: Chapter 18Slide 21 of 42
Example 18-5
1 mol NaC2H3O2
82.0 g NaC2H3O2
mass C2H3O2
-
= 0.300 L
[C2H3O2
-
] = 0.56 M
1 L
0.56 mol
1 mol C2H3O2
-
1 mol NaC2H3O2
 
 = 14 g NaC2H3O2

22. Prentice-Hall General Chemistry: Chapter 18Slide 22 of 42
Six Methods of Preparing Buffer Solutions

23. Prentice-Hall General Chemistry: Chapter 18Slide 23 of 42
Calculating Changes in Buffer Solutions

24. Prentice-Hall General Chemistry: Chapter 18Slide 24 of 42
Buffer Capacity and Range
• Buffer capacity is the amount of acid or base that a
buffer can neutralize before its pH changes
appreciably.
– Maximum buffer capacity exists when [HA] and [A-
]
are large and approximately equal to each other.
• Buffer range is the pH range over which a buffer
effectively neutralizes added acids and bases.
– Practically, range is 2 pH units around pKa

25. Prentice-Hall General Chemistry: Chapter 18Slide 25 of 42
18-3 Acid-Base Indicators
• Color of some substances depends on the pH.
HIn + H2O  In-
+ H3O+
>90% acid form the color appears to be the acid color
>90% base form the color appears to be the base color
Intermediate color is seen in between these two states.
Complete color change occurs over 2 pH units.

26. Prentice-Hall General Chemistry: Chapter 18Slide 26 of 42
Indicator Colors and Ranges

27. Prentice-Hall General Chemistry: Chapter 18Slide 27 of 42
18-4 Neutralization Reactions and
Titration Curves
• Equivalence point:
– The point in the reaction at which both acid and base have been
consumed.
– Neither acid nor base is present in excess.
• End point:
– The point at which the indicator changes color.
• Titrant:
– The known solution added to the solution of unknown
concentration.
• Titration Curve:
– The plot of pH vs. volume.

28. Prentice-Hall General Chemistry: Chapter 18Slide 28 of 42
The millimole
• Typically:
– Volume of titrant added is less than 50 mL.
– Concentration of titrant is less than 1 mol/L.
– Titration uses less than 1/1000 mole of acid and base.
L/1000
mol/1000
=M =
L
mol
mL
mmol
=

29. Prentice-Hall General Chemistry: Chapter 18Slide 29 of 42
Titration of a Strong Acid
with a Strong Base

30. Prentice-Hall General Chemistry: Chapter 18Slide 30 of 42
Titration of a Strong Acid
with a Strong Base
• The pH has a low value at the beginning.
• The pH changes slowly
– until just before the equivalence point.
• The pH rises sharply
– perhaps 6 units per 0.1 mL addition of titrant.
• The pH rises slowly again.
• Any Acid-Base Indicator will do.
– As long as color change occurs between pH 4 and 10.

31. Prentice-Hall General Chemistry: Chapter 18Slide 31 of 42
Titration of a Strong Base
with a Strong Acid

32. Prentice-Hall General Chemistry: Chapter 18Slide 32 of 42
Titration of a Weak Acid
with a Strong Base

33. Prentice-Hall General Chemistry: Chapter 18Slide 33 of 42
Titration of a Weak Acid
with a Strong Base

34. Prentice-Hall General Chemistry: Chapter 18Slide 34 of 42
Titration of a Weak Polyprotic Acid
H3PO4  H2PO4
-
 HPO4
2-
 PO4
3-
NaOHNaOH

35. Prentice-Hall General Chemistry: Chapter 18Slide 35 of 42
18-5 Solutions of Salts of Polyprotic Acids
• The third equivalence point of phosphoric acid can
only be reached in a strongly basic solution.
• The pH of this third equivalence point is not
difficult to caluclate.
– It corresponds to that of Na3PO4 (aq) and PO4
3-
can ionize
only as a base.
PO4
3-
+ H2O → OH-
+ HPO4
2-
Kb = Kw/Ka = 2.410-2

36. Prentice-Hall General Chemistry: Chapter 18Slide 36 of 42
Example 18-9
Kb = 2.410-2
PO4
3-
+ H2O → OH-
+ HPO4
2-
Initial concs. 1.0 M 0 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (1.00 - x) M x M x M
Determining the pH of a Solution Containing the Anion (An-
) of
a Polyprotic Acid.
Sodium phosphate, Na3PO4, is an ingredient of some
preparations used to clean painted walls before they are
repainted. What is the pH of 1.0 M Na3PO4?

37. Prentice-Hall General Chemistry: Chapter 18Slide 37 of 42
Example 18-9
x2
+ 0.024x – 0.024 = 0 x = 0.14 M
pOH = +0.85 pH = 13.15
[OH-
] [HPO4
2-
]
[PO4
3-
]
Kb=
x · x
(1.00 - x)
= = 2.410-2
It is more difficult to calculate the pH values of NaH2PO4 and
Na2HPO4 because two equilibria must be considered
simultaneously.

38. Prentice-Hall General Chemistry: Chapter 18Slide 38 of 42
Concentrated Solutions of
Polyprotic Acids
• For solutions that are reasonably concentrated
(> 0.1 M) the pH values prove to be independent
of solution concentrations.
for H2PO4
-
for HPO4
2-
pH = 0.5 (pKa1 + pKa2) = 0.5 (2.15 + 7.20) = 4.68
pH = 0.5 (pKa1 + pKa2) = 0.5 (7.20 + 12.38) = 9.79

39. Prentice-Hall General Chemistry: Chapter 18Slide 39 of 42
18-6 Acid-Base Equilibrium Calculations:
A Summary
• Determine which species are potentially present in
solution, and how large their concentrations are
likely to be.
• Identify possible reactions between components
and determine their stoichiometry.
• Identify which equilibrium equations apply to the
particular situation and which are most significant.

40. Prentice-Hall General Chemistry: Chapter 18Slide 40 of 42
Focus On Buffers in Blood
CO2(g) + H2O  H2CO3(aq)
H2CO3(aq) + H2O(l)  HCO3
-
(aq)
Ka1 = 4.410-7
pKa1 = 6.4
pH = 7.4 = 6.4 +1.0
pH = pKa1 + log
[H2CO3]
[HCO3
-
]

41. Prentice-Hall General Chemistry: Chapter 18Slide 41 of 42
Buffers in Blood
• 10/1 buffer ratio is somewhat outside maximum
buffer capacity range but…
• The need to neutralize excess acid (lactic) is
generally greater than the need to neutralize excess
base.
• If additional H2CO3 is needed CO2 from the lungs can
be utilized.
• Other components of the blood (proteins and
phosphates) contribute to maintaining blood pH.

42. Prentice-Hall General Chemistry: Chapter 18Slide 42 of 42
Chapter 18 Questions
Develop problem solving skills and base your strategy not
on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who
have been here before.